Question

已知f(x)是定义在R上的函数，对于任意实数a，b都有f(ab)=af(b)+bf(a)，且f(2)=1求f(1\2)的值

Answer 1

f(2) = f(1*2)
= f(2) +2f(1)
f(1) = 0
f(1)= f(2*1/2) = 2f(1/2)+1/2 * f(2)
2f(1/2) + 1/2 = 0
f(1/2) = - 1/4

追问

求f(2^-n)的解析式

追答

f(2^-n)
=f(1/2^n)

f(1/2^-2)= f(1/4) = f(1/2 * 1/2)
= 1/2 * f(1/2) + 1/2 * f( 1/2)
= f(1/2) =- 1/4 =- 2/8

f(2^-3)=f(1/8) = f(1/2 * 1/4)
= 1/2 * f(1/4) + 1/4* f(1/2)
= 1/2 * (-1/4) + 1/4 * (-1/4)
= 3/4 * (-1/4)
=-3/16
f(1/2^-4)=f(1/6)=f(1/4 * 1/4)
= 1/2* f(1/4)
= -1/8 = - 4/32
f(1/2^4) = 1/2 * f(1/8) + 1/8 * f(1/2)
= 1/2* (-3/16 )+ 1/8 *(-1/4)
= - 3/32 - 1/32
= -1/8

...

f(1/2^-n) = - n/ 2^(n+2) 意思： 2的（n+2）次方 分之 （-n）

不清楚再问

Answer 2

由题目知 f（1）=f（1）+f（1）所以f（1）=0.5